Axial dipole field twice equatorial

A short dipole has charges +q and −q separated by 2a, with dipole moment p = 2qa. We find the field at distance r from the centre (r ≫ a), where k = 1/4πε₀.

Field on the axial line

   -q          +q           P
    o----------o-----------x---> 
   -a    0     +a     r

1. P is on the line joining the charges. Distances from +q and −q are (r − a) and (r + a).
2. Field of +q (away from the charge, along p): kq/(r − a)². Field of −q (towards the charge, opposite to p): kq/(r + a)².
3. E_axial = kq [1/(r − a)² − 1/(r + a)²] = kq · 4ar / (r² − a²)².
4. For r ≫ a: E_axial ≈ 4kqa / r³ = 2kp / r³, along p.

Field on the equatorial line

          P
         /|\
      r1/ | \r2
       /  |r \
   -q o---+---o +q

1. P lies on the perpendicular bisector, at a distance r from the centre. Each charge is at distance √(r² + a²).
2. Magnitude of each field: E₁ = E₂ = kq / (r² + a²).
3. Components perpendicular to the dipole axis (along OP) cancel. The components parallel to the axis add up, each with cos θ = a/√(r² + a²), directed from +q towards −q, i.e. opposite to p: E_eq = 2 kq/(r² + a²) · a/√(r² + a²) = 2kqa / (r² + a²)^(3/2) = kp / (r² + a²)^(3/2)
4. For r ≫ a: E_eq ≈ kp / r³, opposite to p.

Ratio

E_axial / E_equatorial = (2kp/r³) / (kp/r³) = 2

Hence the axial field of a short dipole is twice the equatorial field in magnitude (the two have opposite directions relative to p).